# Question d4bd9

Sep 4, 2016

$1.635 \times {10}^{2} \text{ Pa.s or Poiseuille (Pl)}$

#### Explanation:

A bubble when placed in a viscous liquid experiences three forces responsible for its movement:

WEIGHT.
Downwards force due to gravity. It has been ignored as given in the problem.
BUOYANCY.
This is upwards force equal to the weight of the displaced liquid displaced. The bubble expands due to decrease of hydrostatic pressure as the bubble gets closer to the surface. Therefore, this force increases as the bubble approaches the surface. For this problem it has been assumed to be constant.
DRAG.
Downwards force, depends on the Reynolds number of the flow. This force is either dependent linearly, or quadratically, on the velocity of bubble.

When the all three forces are in equilibrium, the velocity of the bubble will not change. It has reached its terminal velocity.

The force of viscosity on a small sphere moving through a viscous fluid is given by the expression:

F_d = 6 π μ R v

where ${F}_{d}$ is drag – acting on the interface between the fluid and the sphere, μ is the dynamic viscosity, $R$ is the radius of the sphere and $v$ is its velocity relative to the fluid.
This is Stokes' law which assumes that flow is laminar, object is a sphere having smooth surface, which moves in homogeneous medium and the sphere and medium don't interfere with each other.

Equating this force with the Buoyancy F_b=4/3πR^3ρg, we get the terminal velocity
Where ρ is the density of the liquid and $g$ is acceleration due to gravity.

6πμRv_t=4/3πR^3ρg
=>v_t=2/9(ρgR^2)/μ
Inserting given values we get
2xx10^-3=2/9xx(1.5xx10^3xx9.81xx(10^-2)^2)/μ#
$\implies \mu = \frac{2}{9} \times \frac{1.5 \times {10}^{3} \times 9.81 \times {\left({10}^{-} 2\right)}^{2}}{2 \times {10}^{-} 3}$
$\implies \mu = 1.635 \times {10}^{2} \text{ Pa.s or Poiseuille (Pl)}$